Lemma 17.27.9. In Lemma 17.27.8 suppose that $\mathcal{O}_2 \to \mathcal{O}'_2$ is surjective with kernel $\mathcal{I} \subset \mathcal{O}_2$ and assume that $\mathcal{O}_1 = \mathcal{O}'_1$. Then there is a canonical exact sequence of $\mathcal{O}'_2$-modules

$\mathcal{I}/\mathcal{I}^2 \longrightarrow \Omega _{\mathcal{O}_2/\mathcal{O}_1} \otimes _{\mathcal{O}_2} \mathcal{O}'_2 \longrightarrow \Omega _{\mathcal{O}'_2/\mathcal{O}_1} \longrightarrow 0$

The leftmost map is characterized by the rule that a local section $f$ of $\mathcal{I}$ maps to $\text{d}f \otimes 1$.

Proof. For a local section $f$ of $\mathcal{I}$ denote $\overline{f}$ the image of $f$ in $\mathcal{I}/\mathcal{I}^2$. To show that the map $\overline{f} \mapsto \text{d}f \otimes 1$ is well defined we just have to check that $\text{d} f_1f_2 \otimes 1 = 0$ if $f_1, f_2$ are local sections of $\mathcal{I}$. And this is clear from the Leibniz rule $\text{d} f_1f_2 \otimes 1 = (f_1 \text{d}f_2 + f_2 \text{d} f_1 )\otimes 1 = \text{d}f_2 \otimes f_1 + \text{d}f_1 \otimes f_2 = 0$. A similar computation show this map is $\mathcal{O}'_2 = \mathcal{O}_2/\mathcal{I}$-linear. The map on the right is the one from Lemma 17.27.8. To see that the sequence is exact, we can check on stalks (Lemma 17.3.1). By Lemma 17.27.7 this follows from Algebra, Lemma 10.131.9. $\square$

Comment #2362 by Dominic Wynter on

That last equation should end in $\mathrm{d}f_2\otimes f_1 + \mathrm{d}f_1\otimes f_2 = 0$, as opposed to $\mathrm{d}f_2\otimes f_1 + \mathrm{d}f_2\otimes f_1 = 0$.

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